Electric field problem

1. Sep 4, 2006

indigojoker

Three point charges are placed on the y-axis. a chage at q at y=a, a charge at -2q at the origin, and a charge q at y=-a. find the magnitude and direction of the electrif field on the y-axis for y>a

I'm not sure where to begin at all. This is what I am thinking, but not completely sure. I could find the electric field the charges but I dont know how to encoporate the fact that there are two other charges that add to the electric field. any suggestions?

2. Sep 4, 2006

Office_Shredder

Staff Emeritus
You can calculate the electric field for each charge individually at a point, then add the three fields together (keeping in mind that opposite direction fields subtract, not add).

3. Sep 5, 2006

indigojoker

k=1/4piε

$$E=k \frac{-q}{4a^2} + k \frac{-2q}{a^2}$$

$$E= k \left( \frac{-q}{4a^2} + k \frac{-8q}{4a^2} \right)$$

$$E=k \left( \frac{-9q}{4a^2} \right)$$

does this look right?

4. Sep 5, 2006

dmoravec

you aren't taking into account the differences in distances to each charge. Lets say you are at a point b along the y axis. then the distance 'r' from the first charge q is b-a. the distance for the one at the orgin (the -2q) is b-0 or b. lastly the distance for the one at -a is b+a. these are the radii you need to plug into the general equation
E=kq/r^2.

5. Sep 5, 2006

indigojoker

the -2q distance is a

the -q distance is 2a

correct?

that is what I have.

6. Sep 5, 2006

dmoravec

the problem stated says 'find the magnitude and direction of the electrif field on the y-axis for y>a'. Therefore the distance is not a set value, but depends on where you want to look (for values greater than a). This is why I used b. b is just an arbitrary y value.

7. Sep 5, 2006

indigojoker

then the solution can be infinite!?

actually, there is a second part to the problem. it asks me to use the binomial expansion to show that very far from the quadrupole the electric field is proportional to y^-4 which i, again, have no clue how to tackle.

8. Sep 5, 2006

dmoravec

well the solution won't be infinite. as the radius goes to infinity the solution goes to zero since the radius is on the bottom.

also not sure about the second part... i might try setting y>>a so that the radii all just go towards y... but that doesn't give you y^-4
or try and use the taylor expansion and throw out small terms...
taylor expansion
(1+x)^r = 1 + rx + 1/2*r(r-1)x^2+1/6*r(r-1)(r-2)x^3...etc

9. Sep 5, 2006

indigojoker

i see what you are saying

b>a

$$E=k \frac{-q}{(a+b)^2} + k \frac{-2q}{b^2}$$

10. Sep 5, 2006

dmoravec

close, but unless I am reading the problem wrong there are 3 charges. And two are positive q (with radii a-b and a+b). The last term is correct though.

11. Sep 5, 2006

indigojoker

$$E=k \frac{q}{(b-a)^2} + k \frac{-2q}{b^2} + k \frac{q}{(a+b)^2}$$

12. Sep 5, 2006

dmoravec

yes, looks good

13. Sep 6, 2006

indigojoker

how do you tell if it goes towards y^-4 or not?

14. Sep 6, 2006

dmoravec

my guess would be to attempt to combine all of the fractions into one and then go from there.

15. Sep 6, 2006

HallsofIvy

Staff Emeritus
Your original problem was "find the magnitude and direction of the electrif field on the y-axis for y>a". None of your formulas include "y"!

From the point at (0,y) to each of the charges, at (0, a), (0, 0), and (0, -a) the distances are |y-a|, |y|, and |y+a| respectively.

16. Sep 6, 2006

dmoravec

17. Sep 6, 2006

indigojoker

the next part of the question asks: Use the binomial expansion to show that very far from the quadrupole, so that y>>a, the electric field is proportional to y^-4. Contrast the behavior to that of the eectric field of a point charge and that of the electric field of an electric dipole.

$$E=k \frac{q}{(y-a)^2} + k \frac{-2q}{y^2} + k \frac{q}{(a+y)^2}$$

when y>>a

$$E=k \frac{q}{y^2} + k \frac{-2q}{y^2} + k \frac{q}{y^2}$$

so E=0

not really sure where to go from here

18. Sep 6, 2006

dmoravec

before assuming that y>>a try moving the fractions into one.
for example, the first term would become:
$$k\frac{q y^2 (a+y)^2}{(y-a)^2y^2(y+a)^2}$$
similarly try this with the other two fractions and then they can be added into one.

19. Sep 7, 2006

indigojoker

$$E= kq \left( \frac{ y^2 (a+y)^2}{(y-a)^2y^2(y+a)^2} - \frac{2 (y-a)^2 (a+y)^2}{(y-a)^2y^2(y+a)^2}+ \frac{ y^2 (a-y)^2}{(y-a)^2y^2(y+a)^2} \right)$$

assuming that y>>a

$$E=kq \frac{y^4-y^4+y^4}{y^6}$$

I get the answer as y^-2

where am I going wrong? and why does the question as me to use the binomial expansion?